Question: Let $f(x)=\dfrac{x+1}{x-3}$ Where does $f$ have critical points? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=-1$ (Choice B) B $x=0$ (Choice C) C $x=3$ (Choice D) D $f$ has no critical points.
A critical point of $f$ is a point in the domain of $f$ where the derivative is either equal to zero or undefined. So in order to find the critical points of $f$, let's find its derivative. $\begin{aligned} f'(x)&=\dfrac{d}{dx}\left[\dfrac{x+1}{x-3}\right] \\\\ &=\dfrac{\dfrac{d}{dx}[x+1](x-3)-(x+1)\dfrac{d}{dx}[x-3]}{(x-3)^2} \\\\ &=\dfrac{(1)(x-3)-(x+1)(1)}{(x-3)^2} \\\\ &=\dfrac{-4}{(x-3)^2} \end{aligned}$ Now let's look for $x$ -values where $f'$ is zero or undefined. $\dfrac{-4}{(x-3)^2}=0$ has no solution, so $f'$ is never equal to zero. $\dfrac{-4}{(x-3)^2}$ is undefined at $x=3$. However, $f(x)=\dfrac{x+1}{x-3}$ is also undefined at $x=3$ so it isn't a critical point. In conclusion, $f$ has no critical points.